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082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

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7.6 The vector product 249

The k component is found by crossing out the row and column containing k and performingthe

calculation

We have

a 1

b 2

−a 2

b 1

Ifa =a 1

i+a 2

j+a 3

kandb =b 1

i+b 2

j+b 3

k,then

∣ i j k ∣∣∣∣∣

a×b=

a 1

a 2

a 3

∣b 1

b 2

b 3

= (a 2

b 3

−a 3

b 2

)i− (a 1

b 3

−a 3

b 1

)j+ (a 1

b 2

−a 2

b 1

)k

Example7.22 Find the vector product ofa = 2i +3j +7k andb = i +2j +k.

Solution The two given vectors arerepresented inthe following determinant:

i jk

237

∣121∣

Evaluating this determinant we obtain

a×b=(3−14)i−(2−7)j+(4−3)k=−11i+5j+k

Youwillfindthat,withpractice,thismethodofevaluatingavectorproductissimpleto

apply.

7.6.2 Applicationsofthevectorproduct

The following threeexamples illustrateapplications of the vector product.

Engineeringapplication7.9

MagneticfluxdensityBandmagneticfieldstrengthH

Itispossibletomodeltheeffectofmagnetismbymeansofavectorfield.Amagnetic

fieldwithmagneticfluxdensityBisavectorfieldwhichisdefinedinrelationtothe

forceitexertsonamovingchargedparticleplacedinthefield.ConsiderFigure7.31.

If a charge q moves with velocity v in a magnetic field B it experiences a force F

given by

F=qv×B

Notethatthisforceisdefinedusingavectorproduct.Theunitofmagneticfluxdensityistheweberpersquaremetre(Wbm

−2 ),ortesla(T).Thedirectionofthisforce

is at right angles to both v and B, in a sense defined by the right-handed screw rule.

Its magnitude, or modulus, is

F =qvBsinθ

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